$A$ circular disc rolls on a horizontal floor without slipping and the centre of the disc moves with a uniform velocity $v$. Which of the following values can the velocity of a point on the rim of the disc have?

The ratio of kinetic energies of two spheres rolling with equal centre of mass velocities is $2 : 1$. If their radii are in the ratio $2 : 1$,then the ratio of their masses will be

$A$ solid sphere is in rolling motion. In rolling motion,a body possesses translational kinetic energy $(K_t)$ as well as rotational kinetic energy $(K_r)$ simultaneously. The ratio $K_t : (K_t + K_r)$ for the sphere is

An object of mass $10 \ kg$ and radius $0.5 \ m$ is rolling without slipping with a velocity of $2 \ m/s$. Its total kinetic energy is $32.8 \ J$. The radius of gyration of the object is .......... $m$.

$A$ sphere is rolling without slipping on a fixed horizontal plane surface. In the figure,$A$ is the point of contact,$B$ is the centre of the sphere,and $C$ is its topmost point. Then:
$(i) \vec{V}_C - \vec{V}_A = 2(\vec{V}_B - \vec{V}_C)$
$(ii) \vec{V}_C - \vec{V}_B = \vec{V}_B - \vec{V}_A$
$(iii) |\vec{V}_C - \vec{V}_A| = 2|\vec{V}_B - \vec{V}_C|$
$(iv) |\vec{V}_C - \vec{V}_A| = 4|\vec{V}_B|$

$A$ force of $49 \ N$ acts tangentially at the highest point of a solid sphere of mass $20 \ kg$,kept on a rough horizontal plane. If the sphere rolls without slipping,then the acceleration of the center of the sphere is (in $m/s^2$)